Is There a Shorter Proof for 0.999... = 1?

[waterchan]

Here is Blizzard's proof that 0.999... = 1

http://www.blizzard.com/press/040401.shtml


My friend however, recently mentioned that:

1/3 = 0.333...

1.3 * 3 = 0.333... * 3

1 = 0.999... also works.


Has she discovered the shortest proof? Or is there something wrong here?

 

[juef]

This is perfectly right. I don't know if there's a "shortest proof", but your friend's certainly is short.

I'm somehow a newbie here, but I used to visit Tom's Hardware Guide's forums, and there's an interesting topic with... various "opinions" on the subject on it. Check it out http://www.community.tomshardware.com/forum/showflat.m?Cat=&Board=other_poll&Number=167099&page=0&view=collapsed&sb=5

 

[BenGoodchild]

There is, ofcourse,

1/9 = 0.111...
multiply be nine
1 = 0.999...

 

[Integral]

This is perfectly right. I don't know if there's a "shortest proof", but your friend's certainly is short.

I'm somehow a newbie here, but I used to visit Tom's Hardware Guide's forums, and there's an interesting topic with... various "opinions" on the subject on it. Check it out http://www.community.tomshardware.com/forum/showflat.m?Cat=&Board=other_poll&Number=167099&page=0&view=collapsed&sb=5

I made a few contribution to that thread. SilverPig started that thread to see if the response was significantly different from a similar thread in the HiTech Forum at anantech, Which I also participated in.

It was very disappointing that the majority of members of Anantech and Toms felt that it was not true.

 

[matt grime]

That isn't a proof. Why is arithmetic deinfed on infinitely long decimals? In short, you're confusing real numbers with their representations as decimals. The fact that they are equal is immediate from the definition of the real numbers, not that anyone who thinks they're different even knows what the real numbers are.

 

[BenGoodchild]

Matt - don't become worried - it isn't meant to be a proof - it is just a cunning trick used by high school mathematics teachers to trick their students and to make them think.

No one is really saying that 0.9 recurring equals 1

Regards

Ben

 

[matt grime]

Oh great; the cranks come out of the wood work. Can we have an instant ban for anyone who, despite the explanation to the contrary being in the thread, asserts that they are not equal?

Apologies if that's a typo and you clarifyting no one *denies* that they are equalivalent as representations of real numbers, or if | misunderstand and you are trying to differentiate between representations of real numbers and the numbers themselves, but I doubt that is your intention.

 

[Integral]

I consider this sort of algebraic manipulation more of a demonstration then a proof. It is a valid demonstration of a mathematical fact, but not a proof.

 

[Zurtex]

Matt - don't become worried - it isn't meant to be a proof - it is just a cunning trick used by high school mathematics teachers to trick their students and to make them think.

No one is really saying that 0.9 recurring equals 1

Regards

Ben

I'm saying 0.9 recurring equals 1 and so will any mathematical approch on it.

 

[arildno]

Now, I don't know if I qualify as one of Zurtex' approch's (sounds like some lumbering, prehistoric animal to me), but I agree with his view as well.

 

[maverickmathematics]

I consider this sort of algebraic manipulation more of a demonstration then a proof. It is a valid demonstration of a mathematical fact, but not a proof.

I'm saying 0.9 recurring equals 1 and so will any mathematical approch on it.

I think BENGOODCHILD was making the point that the value of 0.999... as a number is not 1.

It is true that 0.999... comes from the formula for the series 9/10 + ;9/10^2 + 9/10^3,

therefore the limit of the series is infact 0.999... and therefore one but the value is different.

Unless we want to start the whole debate on infinity and what happens at the last 9 etc - well there i no last 9 becasue 0.99...is non-equatable. Okay?!


-M

 

[arildno]

Yes it it is, maverick (goodchild?). Learn about how real numbers are defined as equivalence classes on the set of (increasing, bounded) sequences of rationals.

 

[matt grime]

Who would have to talk about infinity and the last 9? You, Maverick? Only those who don't understand mathematics would cite that. Indeed there is no reason to invoke infinity at all, indeed the appearance of any infinty is only a short hand fomr something to do with finite things and we need not ever mention it. Now, as I'm apparently not in a charitable mood, can the cranks go away?

All refutations of this fact arise from not understanding maths - the definitions are straight forward, though hard to visualize perhaps, but in the completion of Q (ie R) those are the same number. Fin. Just as 1/2 and 2/4 are the same rational number.

 

[HallsofIvy]

I think BenGoodchild was making a joke.

I hope maverickmathematics was also (look at the user name!).

 

[BenGoodchild]

we're studying at Trinity College Cambridge- look at the notes mav posted about number theory, so yes I'm messing you guys around.

and maverick is a long time friend of mine so there you go

regards,

Ben

 

[arildno]

Level of jocular funniness:
Harrumph, heh-heh

 

[matt grime]

I think BENGOODCHILD was making the point that the value of 0.999... as a number is not 1.

It is true that 0.999... comes from the formula for the series 9/10 + ;9/10^2 + 9/10^3,

therefore the limit of the series is infact 0.999... and therefore one but the value is different.

Unless we want to start the whole debate on infinity and what happens at the last 9 etc - well there i no last 9 becasue 0.99...is non-equatable. Okay?!


-M

I note you've added things since your orignal post.

Funny? Nurse, my sides have split.

 

[BenGoodchild]

you guys are just plain boring - you'll get all worked up if I tell you that 2+2=5 and start crying...

 

[Zurtex]

we're studying at Trinity College Cambridge- look at the notes mav posted about number theory, so yes I'm messing you guys around.

and maverick is a long time friend of mine so there you go

regards,

Ben

I'm studying number theory at UMIST and I'm sure half the people in my class wouldn't know that 0.999... = 1 :grumpy:

 

[arildno]

And, from BenGoodchild's second post, I don't think he knows either.
This damage control action he's undertaken afterwards is unconvincing.

 

[BenGoodchild]

Give me a couple of hours and I'll bring back evidence!

 

[kreil]

I'm thoroughly convinced that 0.9r=1. I have been reading the posts on the hardware forum and here all day during study halls, and when my calculus class finally came I asked my teacher what she thought. Her reply was that it did, but one of my friends wasn't convinced. I argued with him about it on the way to physics. Knowing that my physics teacher is educated in math, I said we would ask him. This is how the conversation went:

"Mr. H, does point 9 repeating equal 1?"
"No"
"Yes it does, we have been arguing about it. Number theory says it does"
"Number theory was created by a bunch of mathematicians that don't know anything about the real world, that's why it is called number theory and not number really"

I walked away, angered and surprised at his ignorance. I never got a chance to show him a proof, which I am hoping will convince him.


Why is this fact so hard for people to accept?

 

[BenGoodchild]

Because to most people it appears as though you are saying somthing equivalent to

0.99999999999999999999999 = 1

They do not understand the actualy CONCEPT of number and to a certain extent it appears as though you are saying that the little bit that one expects should be inbetween the 0.9r and 1 does not exist - people cannot accept it.

For them 0.999999999999999 + 0.000000000000001 = 1

It is the idea of a repeating decimal people cannot fathom

 

[tongos]

how do we know that 1/3=0.33333... by this, we need to look at the geometric series to prove such.

 

[waterchan]

If 1/3 cannot be defined as 0.333...

Then wouldn't that imply that it is illegal to even write the equation 0.999... = 1?

 

[arildno]

1/3 isn't defined as 0.3333..
where did you get that idea from?

 

[tongos]

1/3 is too. simple division leads me to believe that.
this is pointless.

 

[arildno]

1/3 is too. simple division leads me to believe that.
this is pointless.

Learn what a DEFINITION is before you speak.

 

[tongos]

its for a different post. i wasnt talking about "is defined as". 1/3=0.333333... by division.

 

[BenGoodchild]

Thats my post on it as promised

Yes, 0.9 recurring is equal to 1.

1/3 = 0.333...
x3
1 = 0.99999...

That is perfectly fine.

Another way to look at it is that 0.9 recurring is 9/10 + 9/100 + 9/1000 + ... and then use the formula for the sum of a geometric series. It's a really nice bit of maths, so in case you don't know it I'll quickly go through it:

q: what is 1/5 + 1/25 + 1/125 + ... + 1/(5^10) ?
a: well, we know it is a number (i.e. this sum is finite as there are only finitely many terms), so call it S (for "sum").
Then S/5 = 1/25 + ... + 1/(5^10) + 1/(5^11), so
S-S/5 = 1/5 - 1/(5^11), and hence S = (5/4)*(1/5 - 1/(5^11), which after simplifying becomes (1-1/(5^10))/4.
There is a formula for this, but the instructive thing is to remember how to derive it, as above. It's not hard at all.

The infinite case is a bit trickier. How do we know that 1/5 + 1/25 + 1/125 + ... is actually a number, i.e. that it isn't infinite? This might seem unimportant, but consider instead the following situation:
S=1+1/2+1/3+1/4+1/5+... (this is called the harmonic series, and is very interesting). It turns out that this tends to infinity, so it is a logical fallacy to say "call this number S" and then use it as if it were a real number. You can get into all kinds of problems (and people did) by making this mistake.
It turns out that geometric series (i.e. those in which each term is a fixed multiple of the term before) always converge which the ratio is <1. The way to see this is to look at the first N terms, and use what we did above. So 1/5 + ... + 1/(5^N) = (1-1/(5^N))/4 as above, and this always less than 1/4, so 1/5 + 1/25 + ... doesn't tend to infinity. Exactly the same thing works for a+ar+ar^2+ar^3+... always converges when -1<r<1 (in the negative case you have to check that the sum doesn't tend to minus infinity either).
Now we know that, we can do the same trick as above:
S=9/10+9/100+... so S/10=9/100+9/1000+... and hence
(1-1/10)S=9/10, i.e. S=1.

I remember quite well being confused by this when my teacher mentioned it first. Here's a question to ask your self: what is 1-0.9999... ? It is clearly not negative. And it's less than 1/10, as 0.9999... is greater than 0.9. And it's less that 1/100, as 0.9999... is greater than 0.99 = 1-1/100. In fact, for any positive integer n you care to name, it is less than 1/(10^n), as 0.9999... is greater than 0.(n 9s). What non-negative number is less than 1/(10^n) for all n? Well, it can only be 0.

The maths of the real numbers (i.e. anything with a decimal expansion, so that includes integers, rational numbers (p/q), solutions of equations (root 2, sqrt(1+sqrt(2)), and even numbers that aren't roots of (polynomial) equations (pi, e, uncountably many others)) is very interesting. It turns out that this concrete construction, via infinite decimals, is not the most useful one. It makes it hard to prove things. There are three other characterisations of the real numbers, which are all equivalent.
(a)any increasing sequence which is bounded above (i.e. doesn't tend to infinity) tends to a limit. (monotone sequences axiom (monotone means strictly increasing or strictly decreasing))
(b)any non-empty set of real number which is bounded above has a least upper bound. (least upper bound axiom)
(c)an infinite number of points in a interval of finite length must have a subsequence which tends to a limit. (Bolzano-Weierstrass axiom) (strictly, the interval must contain it's end-points)

So to prove things in the real numbers, you choose on of the above axioms and use that and all the facts you know about the rationals (you're working in the smallest field containing the rationals such that your axiom is true). It is something you have to learn in the first year of a maths degree to prove that each axiom is equivalent. That would take too long for me to explain now, but it is worth seeing why these don't work in the rationals.
The thing that's hard to grasp the first time you see this is that when we say "tends to a limit" we mean "there is a point in the field we are considering which this sequences tends to".

(a)take succeedingly better approximations for pi. so 0,1,2,3,3.1,3.14,3.141,etc... This is increasing, bounded above (by 4, say), but if it did tend to a limit then that limit would have to be pi, and pi isn't a rational number.
(b)just take the set of all points I outlined above.
(c)again, the set above works, as it is contained in [0,4].

Thats my post on it as promised

 

[arildno]

its for a different post. i wasnt talking about "is defined as". 1/3=0.333333... by division.

Sorry about that..

 

[arildno]

Since it now is evident that you tried to make a joke last time, Ben, I wish you better luck with your next one..

 

[maverickmathematics]

And now everybody goes quiet when they can't take the michael out of us - nice post Ben and let's just kick it !

-M

 

[Chronos0]

Ok, here's a situation:

You're playing a game, and your score is measured by your hit percentage. You miss the first shot, but then make an infinite amount of hits. What is your score?

Well, your percentage would be (INF - 1) / INF, which equals the closest number to and below 1 (its 1 - (1/INF)), aka 0.99999... etc. By what they say 0.99999... etc. equals 100%, but how can that be if you missed? Does the first shot not count because of all the hits? Are you going to say that 1 / INF is equal to 0? But each hit is worth 1 / INF percent, so then the total percentage would then equal 0 as well...

What, in fact, would be your score?

 

[matt grime]

you guys are just plain boring - you'll get all worked up if I tell you that 2+2=5 and start crying...

search the forums for threads with 0.99 recurring arguments in them then get back to us.